Unformatted text preview: MATH 231 / Spring 2008 Name: Key 1. (4 points) Convert the point (x, y, z) =
1 1 1 2, 2, 2 April 2, 2008 into cylindrical and spherical. Solution: For cylindrical, we're already part of the way there: we know z. Thus, we only need 2 1 1 2 compute r and . We have an equation for r; namely, r2 = 1 + 2 = 1 , so r = 2 . Furthermore, 2 2 we know that = tan1 (1) = /4.
1 1 Now for spherical. For spherical, we know that 2 = x2 + y 2 + z 2 = r2 + z 2 = 2 + 2 = 1. Also, we r already computed in our cylindrical, so we only need to compute . We know that = tan1 ( z ) = 1 tan (1) = /4. f (x, y, z) dV as an iterated integral in 6 different ways, where E 2. (6 points) Express the integral E is the solid bounded by the surfaces x = 2, y = 2, z = 0, and x + y  2z = 2. Solution: First, let's graph the region that we're looking at: 1.0 0.5 2.0 0.0 0.0 0.5 1.0 0.5 1.5 2.0 0.0 1.0 1.5 Note that I have set this up in a rather nonstandard orientation for better viewing. The origin of this graph is located at the bottom left corner of the box. As such, the positive y axis moves up and right, whereas the positive x moves down and right. From here we can pretty much read off what our answers will be:
2 2 (x+y2)/2 2 2 (x+y2)/2 2 y/2 2 f (x, y, z) dz dy dx, 0 2y 0 f (x, y, z) dz dx dy, 0 0 f (x, y, z) dx dz dy, 0 2x 0 2y+2z 1 2 2 2 x/2 2 1 2 2 f (x, y, z) dx dy dz, 0 0 f (x, y, z) dy dz dx, and 0 2z 2x+2z f (x, y, z) dy dx dz. 0 2z 2y+2z 2x+2z 1 ...
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This homework help was uploaded on 04/17/2008 for the course MATH 231 taught by Professor Bociu during the Spring '08 term at UVA.
 Spring '08
 BOCIU
 Math, Calculus

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